package org.usmile.algorithms.leetcode.middle;

/**
 * 200. 岛屿数量
 *
 * 给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
 *
 * 岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
 *
 * 此外，你可以假设该网格的四条边均被水包围。
 *
 * 示例 1：
 *
 * 输入：grid = [
 *   ["1","1","1","1","0"],
 *   ["1","1","0","1","0"],
 *   ["1","1","0","0","0"],
 *   ["0","0","0","0","0"]
 * ]
 * 输出：1
 * 示例 2：
 *
 * 输入：grid = [
 *   ["1","1","0","0","0"],
 *   ["1","1","0","0","0"],
 *   ["0","0","1","0","0"],
 *   ["0","0","0","1","1"]
 * ]
 * 输出：3
 *
 * 提示：
 *
 * m == grid.length
 * n == grid[i].length
 * 1 <= m, n <= 300
 * grid[i][j] 的值为 '0' 或 '1'
 */
public class _0200 {
}


class _0200_Solution {
    private static final int[][] DIRECTIONS = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

    private int rows;
    private int cols;
    private boolean[][] visited;
    private char[][] grid;

    public int numIslands(char[][] grid) {
        this.grid = grid;
        this.rows = grid.length;
        this.cols = grid[0].length;
        visited = new boolean[rows][cols];

        int count = 0;
        for (int row = 0; row < rows; row++) {
            for (int col = 0; col < cols; col++) {
                if (grid[row][col] == '1' && !visited[row][col]) {
                    visited[row][col] = true;
                    count++;
                    dfs(row, col);
                }
            }
        }

        return count;
    }

    private void dfs(int row, int col) {
        for (int[] direction : DIRECTIONS) {
            int nextRow = row + direction[0];
            int nextCol = col + direction[1];
            if (nextRow >= 0 && nextRow < rows && nextCol >= 0 && nextCol < cols) {
                if (grid[nextRow][nextCol] == '1' && !visited[nextRow][nextCol]) {
                    visited[nextRow][nextCol] = true;
                    dfs(nextRow, nextCol);
                }
            }
        }
    }
}